Problem 27 - Differential Equations and Linear Algebra

Question

Determine three linearly independent solutions to the given differential equation of the form \(y(x)=e^{r x},\) and thereby determine the general solution to the differential equation. $$y^{\prime \prime \prime}-3 y^{\prime \prime}-y^{\prime}+3 y=0$$

Step-by-Step Solution

Answer

The general solution to the given differential equation \(y^{\prime\prime\prime}-3y^{\prime\prime}-y^{\prime}+3y=0\) is: \( y(x) = C_1 e^{-x} + C_2 e^{x} + C_3 e^{3x} \) where \(C_1, C_2,\) and \(C_3\) are constants.

Step 1: Form the Characteristic Equation

Replace the y and its derivatives with the proposed exponential function type and its derivatives, then simplify: \( y(x) = e^{r x} \) \( y'(x) = r e^{r x} \) \( y''(x) = r^2 e^{r x} \) \( y'''(x) = r^3 e^{r x} \) Substitute these expressions into the given differential equation and factor out \( e^{r x} \): \( (r^3 - 3r^2 - r +3)e^{r x} = 0 \) Since \( e^{r x} \) is never zero, we can divide both sides by \( e^{r x} \) and obtain the characteristic equation: \( r^3 - 3r^2 - r + 3 = 0 \)

Step 2: Find the Roots

We need to find the roots of the characteristic equation: \( r^3 - 3r^2 - r + 3 = 0 \) By trying values of r or using numerical methods, we find three roots: r₁ = -1, r₂ = 1, r₃ = 3 These roots correspond to exponentials in the form \(y(x) = e^{r x}\), so we have: \( y_1(x) = e^{-x} \) \( y_2(x) = e^{x} \) \( y_3(x) = e^{3x} \)

Step 3: Write the General Solution

Now that we have found three linearly independent solutions, we can write the general solution for the given differential equation as a linear combination of these solutions: \( y(x) = C_1 e^{-x} + C_2 e^{x} + C_3 e^{3x} \) Here, \( C_1, C_2, \) and \( C_3 \) are constants which may be determined by initial or boundary conditions. Without any additional information, this is the general solution to the given differential equation.

Key Concepts

Linear Independence

One of the key steps in solving differential equations, particularly of higher order, is ensuring that the solutions are linearly independent. Linear independence is a crucial concept because it enables us to build the general solution to a differential equation. Two or more functions are said to be linearly independent if no function in the set can be written as a linear combination of the others. For example, if we have solutions \( y_1(x), y_2(x), \) and \( y_3(x), \) they are linearly independent if none of these solutions can be expressed as a combination of the others like \[ ay_1(x) + by_2(x) + cy_3(x) = 0 \] where \( a, b, \) and \( c \) are constants, without all of them being zero.When the solutions to the differential equation are linearly independent, they can form the basis for the general solution. This allows for the combination of these functions with arbitrary constants, creating a family of solutions that satisfies the differential equation under various initial conditions. In our case, the solutions found are \( e^{-x}, e^{x}, \) and \( e^{3x} \), which are linearly independent. Their independence is inferred by the fact that each function grows or decays at different rates, making them unique and not expressible as a simple combination of each other.

Characteristic Equation

The characteristic equation is a fundamental tool in solving linear differential equations. It is derived from substituting a trial solution, usually of the form \( y(x) = e^{rx} \), into the differential equation and simplifying it to obtain an algebraic equation in terms of \( r \). This transformed equation is known as the characteristic equation.For the given differential equation, we first compute the derivatives of the trial solution and substitute:

• \( y'(x) = re^{rx} \)
• \( y''(x) = r^2e^{rx} \)
• \( y'''(x) = r^3e^{rx} \)

It leads us to the characteristic equation after factoring out \( e^{rx} \):\[ r^3 - 3r^2 - r + 3 = 0 \]The roots of this characteristic equation provide the values of \( r \) that correspond to the exponential solutions \( y(x) = e^{rx} \). Solving this equation helps in finding the independent solutions necessary for constructing the general solution to the differential equation.

Exponential Solutions

Exponential functions are frequently used as trial solutions for differential equations, especially linear ones with constant coefficients. The form \( y(x) = e^{rx} \) is chosen because its derivatives maintain the exponential function's form, simplifying the substitution process when forming the characteristic equation. The beauty of exponential solutions lies in their simplicity and generality. Once the roots of the characteristic equation are determined, the exponential solutions \( y(x) = e^{rx} \) can be directly used. Each root \( r \) represents a distinct solution, and these solutions can be combined to form a general solution.In this exercise, the roots found were \( r = -1, 1, \) and \( 3 \), leading to the corresponding exponential solutions:

• \( y_1(x) = e^{-x} \)
• \( y_2(x) = e^{x} \)
• \( y_3(x) = e^{3x} \)

These solutions are linearly independent and can be expressed in their general form as a linear combination: \[ y(x) = C_1 e^{-x} + C_2 e^{x} + C_3 e^{3x} \]Exponential solutions provide a straightforward way to describe a whole set of solutions to the differential equation, adaptable to boundary or initial conditions by adjusting the constants \( C_1, C_2, \) and \( C_3 \). Understanding these concepts allows for powerful solution strategies in differential equations.